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%I #11 Nov 25 2019 00:58:30
%S 1,2,3,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12,13,13,14,14,15,16,16,17,18,
%T 18,19
%N Number of weighings needed to find lighter coins among n coins.
%C a(n) is the minimal worst-case number of weighings needed by an algorithm to sort a set of n coins of two possible weights into heavy vs. light coins by using a balance; additional known good (heavy, say) coins are available (esp. to distinguish "all heavy" from "all light").
%C It is known that ceiling(n*log_3(2) + log_3(135/128)) <= a(n) <= ceiling(7n/11) for all n except for n=3. This implies 19 <= a(30) <= 20 and in fact a(n) = ceiling(7n/11) for several n <= 187, esp. for all n with n <= 50 except n=3, n=30, n=41, n=49.
%H An-Ping Li, <a href="http://arxiv.org/abs/0902.0841">A note on the counterfeit coins problem</a>, arXiv:0902.0841v8 [math.CO]
%e For n=1, compare the given coin A with a known heavy coin X. If A < X, then A is a light coin; if A=X, then A is a heavy coin; the outcome A > X is not possible. Since one weighing was needed, we have a(1)=1.
%e For n=3, to sort coins A,B,C, one optimal algorithm is: First compare A:B. If A < B, we know that A is light and B is heavy and can find out about C by comparing, e.g., B:C in a second weighing. The case A > B is symmetric. If, however, A=B, compare A:C. If A < C, we know that A,B are light and C is heavy and vice versa for A > C. The worst case is A=C, which requires a third weighing, e.g., A:X, against a known heavy coin X. Since no algorithm exists that never uses more than 2 weighings, we have a(3) = 3.
%Y Cf. A156301. - _Jonathan Vos Post_, May 18 2009
%K nonn,hard,more
%O 1,2
%A _Hagen von Eitzen_, May 16 2009
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